A 75 kVA three-phase transformer steps 480 volts down to 208Y/120 volts and is marked with a nameplate impedance of 2.5 percent. Its primary full-load current is 90 amperes and its secondary full-load current is 208 amperes. Available fault current at the secondary terminals is approximated by dividing the secondary full-load current by the per-unit impedance. What is the approximate available fault current at those terminals?
- AAbout 14,400 amperes
This comes from computing the secondary current without the square root of three, treating a three-phase transformer as though it were single phase, then dividing by the impedance.
- BAbout 3,600 amperes
This divides the primary full-load current by the impedance. Fault current at the secondary terminals must be based on the secondary current.
- About 8,300 amperes
- DAbout 830 amperes
This divides by 25 percent rather than 2.5 percent, a decimal slip that understates the fault current by a factor of ten and would justify badly underrated equipment.
Why C is correct
The approximation divides secondary full-load current by per-unit impedance: 208 amperes divided by 0.025 gives about 8,320 amperes. Lower nameplate impedance means a stiffer transformer and higher fault current, which is why a low-impedance unit can push equipment past its rating even on a modest kVA.
What this question is testing
This tests whether you can carry a two-step calculation and whether you keep primary and secondary quantities straight. The distractors are built from the three errors that actually occur: dropping the square root of three, using the wrong side of the transformer, and misplacing the decimal on the impedance.
On the job
This calculation is the first thing done when a panel or a control cabinet has to be matched to a service. Utilities publish available fault current at the service point, and a transformer inside the building sets it again for everything downstream. The number falls off quickly along a feeder, which is why a panel 200 feet away can use lower rated devices than one bolted to the transformer. When equipment cannot be rated high enough, the fix is a current-limiting device ahead of it, and that solution has to be engineered rather than guessed.
Memory technique
Full-load current divided by per-unit impedance gives the fault.
Exam tip
Write the impedance as a decimal before you divide. Most errors on this calculation are decimal errors, not conceptual ones.
Where to look it up
Article 110 Part I, the interrupting rating section and the field marking of available fault current a few sections later. The transformer data comes from the equipment nameplate rather than from a Code table.